Integrand size = 18, antiderivative size = 76 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {1}{2} a^2 A x^2+\frac {b (A b+2 a B) x^{2 (1+n)}}{2 (1+n)}+\frac {a (2 A b+a B) x^{2+n}}{2+n}+\frac {b^2 B x^{2+3 n}}{2+3 n} \] Output:
1/2*a^2*A*x^2+b*(A*b+2*B*a)*x^(2+2*n)/(2+2*n)+a*(2*A*b+B*a)*x^(2+n)/(2+n)+ b^2*B*x^(2+3*n)/(2+3*n)
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {1}{2} x^2 \left (a^2 A+\frac {2 a (2 A b+a B) x^n}{2+n}+\frac {b (A b+2 a B) x^{2 n}}{1+n}+\frac {2 b^2 B x^{3 n}}{2+3 n}\right ) \] Input:
Integrate[x*(a + b*x^n)^2*(A + B*x^n),x]
Output:
(x^2*(a^2*A + (2*a*(2*A*b + a*B)*x^n)/(2 + n) + (b*(A*b + 2*a*B)*x^(2*n))/ (1 + n) + (2*b^2*B*x^(3*n))/(2 + 3*n)))/2
Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {950, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (a^2 A x+a x^{n+1} (a B+2 A b)+b x^{2 n+1} (2 a B+A b)+b^2 B x^{3 n+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} a^2 A x^2+\frac {b x^{2 (n+1)} (2 a B+A b)}{2 (n+1)}+\frac {a x^{n+2} (a B+2 A b)}{n+2}+\frac {b^2 B x^{3 n+2}}{3 n+2}\) |
Input:
Int[x*(a + b*x^n)^2*(A + B*x^n),x]
Output:
(a^2*A*x^2)/2 + (b*(A*b + 2*a*B)*x^(2*(1 + n)))/(2*(1 + n)) + (a*(2*A*b + a*B)*x^(2 + n))/(2 + n) + (b^2*B*x^(2 + 3*n))/(2 + 3*n)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^ n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGt Q[p, 0] && IGtQ[q, 0]
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {a \left (2 A b +B a \right ) x^{2} x^{n}}{2+n}+\frac {b^{2} B \,x^{2} x^{3 n}}{2+3 n}+\frac {a^{2} A \,x^{2}}{2}+\frac {b \left (A b +2 B a \right ) x^{2} x^{2 n}}{2+2 n}\) | \(76\) |
norman | \(\frac {a \left (2 A b +B a \right ) x^{2} {\mathrm e}^{n \ln \left (x \right )}}{2+n}+\frac {b^{2} B \,x^{2} {\mathrm e}^{3 n \ln \left (x \right )}}{2+3 n}+\frac {a^{2} A \,x^{2}}{2}+\frac {b \left (A b +2 B a \right ) x^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{2+2 n}\) | \(82\) |
parallelrisch | \(\frac {2 B \,x^{2} x^{3 n} b^{2} n^{2}+3 A \,x^{2} x^{2 n} b^{2} n^{2}+6 B \,x^{2} x^{3 n} b^{2} n +6 B \,x^{2} x^{2 n} a b \,n^{2}+8 A \,x^{2} x^{2 n} b^{2} n +12 A \,x^{2} x^{n} a b \,n^{2}+3 A \,x^{2} a^{2} n^{3}+4 b^{2} B \,x^{2} x^{3 n}+16 B \,x^{2} x^{2 n} a b n +6 B \,x^{2} x^{n} a^{2} n^{2}+4 A \,x^{2} x^{2 n} b^{2}+20 A \,x^{2} x^{n} a b n +11 A \,x^{2} a^{2} n^{2}+8 B \,x^{2} x^{2 n} a b +10 B \,x^{2} x^{n} a^{2} n +8 A \,x^{2} x^{n} a b +12 A \,x^{2} a^{2} n +4 B \,x^{2} x^{n} a^{2}+4 a^{2} A \,x^{2}}{2 \left (2+n \right ) \left (2+3 n \right ) \left (1+n \right )}\) | \(276\) |
orering | \(\frac {3 x^{2} \left (2 n^{2}+9 n +5\right ) \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{4 \left (3 n^{2}+8 n +4\right )}-\frac {\left (11 n +7\right ) x^{2} \left (\left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )+2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n +\left (a +b \,x^{n}\right )^{2} B \,x^{n} n \right )}{4 \left (3 n^{2}+8 n +4\right )}+\frac {x^{3} \left (1+3 n \right ) \left (\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{x}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{x}+\frac {2 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{x}+\frac {4 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{x}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{x}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{x}\right )}{6 n^{3}+22 n^{2}+24 n +8}-\frac {x^{4} \left (-\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{x^{2}}-\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{x^{2}}+\frac {6 b^{2} x^{2 n} n^{3} \left (A +B \,x^{n}\right )}{x^{2}}+\frac {6 b^{2} x^{3 n} n^{3} B}{x^{2}}+\frac {12 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{3} b}{x^{2}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{3}}{x^{2}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{3}}{x^{2}}\right )}{4 \left (3 n^{3}+11 n^{2}+12 n +4\right )}\) | \(456\) |
Input:
int(x*(a+b*x^n)^2*(A+B*x^n),x,method=_RETURNVERBOSE)
Output:
a*(2*A*b+B*a)/(2+n)*x^2*x^n+b^2*B/(2+3*n)*x^2*(x^n)^3+1/2*a^2*A*x^2+1/2*b* (A*b+2*B*a)/(1+n)*x^2*(x^n)^2
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (72) = 144\).
Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.49 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2 \, {\left (B b^{2} n^{2} + 3 \, B b^{2} n + 2 \, B b^{2}\right )} x^{2} x^{3 \, n} + {\left (8 \, B a b + 4 \, A b^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} n^{2} + 8 \, {\left (2 \, B a b + A b^{2}\right )} n\right )} x^{2} x^{2 \, n} + 2 \, {\left (2 \, B a^{2} + 4 \, A a b + 3 \, {\left (B a^{2} + 2 \, A a b\right )} n^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} n\right )} x^{2} x^{n} + {\left (3 \, A a^{2} n^{3} + 11 \, A a^{2} n^{2} + 12 \, A a^{2} n + 4 \, A a^{2}\right )} x^{2}}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \] Input:
integrate(x*(a+b*x^n)^2*(A+B*x^n),x, algorithm="fricas")
Output:
1/2*(2*(B*b^2*n^2 + 3*B*b^2*n + 2*B*b^2)*x^2*x^(3*n) + (8*B*a*b + 4*A*b^2 + 3*(2*B*a*b + A*b^2)*n^2 + 8*(2*B*a*b + A*b^2)*n)*x^2*x^(2*n) + 2*(2*B*a^ 2 + 4*A*a*b + 3*(B*a^2 + 2*A*a*b)*n^2 + 5*(B*a^2 + 2*A*a*b)*n)*x^2*x^n + ( 3*A*a^2*n^3 + 11*A*a^2*n^2 + 12*A*a^2*n + 4*A*a^2)*x^2)/(3*n^3 + 11*n^2 + 12*n + 4)
Leaf count of result is larger than twice the leaf count of optimal. 767 vs. \(2 (66) = 132\).
Time = 0.53 (sec) , antiderivative size = 767, normalized size of antiderivative = 10.09 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\begin {cases} \frac {A a^{2} x^{2}}{2} + 2 A a b \log {\left (x \right )} - \frac {A b^{2}}{2 x^{2}} + B a^{2} \log {\left (x \right )} - \frac {B a b}{x^{2}} - \frac {B b^{2}}{4 x^{4}} & \text {for}\: n = -2 \\\frac {A a^{2} x^{2}}{2} + 2 A a b x + A b^{2} \log {\left (x \right )} + B a^{2} x + 2 B a b \log {\left (x \right )} - \frac {B b^{2}}{x} & \text {for}\: n = -1 \\\frac {A a^{2} x^{2}}{2} + \frac {3 A a b x^{\frac {4}{3}}}{2} + \frac {3 A b^{2} x^{\frac {2}{3}}}{2} + \frac {3 B a^{2} x^{\frac {4}{3}}}{4} + 3 B a b x^{\frac {2}{3}} + B b^{2} \log {\left (x \right )} & \text {for}\: n = - \frac {2}{3} \\\frac {3 A a^{2} n^{3} x^{2}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {11 A a^{2} n^{2} x^{2}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {12 A a^{2} n x^{2}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {4 A a^{2} x^{2}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {12 A a b n^{2} x^{2} x^{n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {20 A a b n x^{2} x^{n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {8 A a b x^{2} x^{n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {3 A b^{2} n^{2} x^{2} x^{2 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {8 A b^{2} n x^{2} x^{2 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {4 A b^{2} x^{2} x^{2 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {6 B a^{2} n^{2} x^{2} x^{n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {10 B a^{2} n x^{2} x^{n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {4 B a^{2} x^{2} x^{n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {6 B a b n^{2} x^{2} x^{2 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {16 B a b n x^{2} x^{2 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {8 B a b x^{2} x^{2 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {2 B b^{2} n^{2} x^{2} x^{3 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {6 B b^{2} n x^{2} x^{3 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} + \frac {4 B b^{2} x^{2} x^{3 n}}{6 n^{3} + 22 n^{2} + 24 n + 8} & \text {otherwise} \end {cases} \] Input:
integrate(x*(a+b*x**n)**2*(A+B*x**n),x)
Output:
Piecewise((A*a**2*x**2/2 + 2*A*a*b*log(x) - A*b**2/(2*x**2) + B*a**2*log(x ) - B*a*b/x**2 - B*b**2/(4*x**4), Eq(n, -2)), (A*a**2*x**2/2 + 2*A*a*b*x + A*b**2*log(x) + B*a**2*x + 2*B*a*b*log(x) - B*b**2/x, Eq(n, -1)), (A*a**2 *x**2/2 + 3*A*a*b*x**(4/3)/2 + 3*A*b**2*x**(2/3)/2 + 3*B*a**2*x**(4/3)/4 + 3*B*a*b*x**(2/3) + B*b**2*log(x), Eq(n, -2/3)), (3*A*a**2*n**3*x**2/(6*n* *3 + 22*n**2 + 24*n + 8) + 11*A*a**2*n**2*x**2/(6*n**3 + 22*n**2 + 24*n + 8) + 12*A*a**2*n*x**2/(6*n**3 + 22*n**2 + 24*n + 8) + 4*A*a**2*x**2/(6*n** 3 + 22*n**2 + 24*n + 8) + 12*A*a*b*n**2*x**2*x**n/(6*n**3 + 22*n**2 + 24*n + 8) + 20*A*a*b*n*x**2*x**n/(6*n**3 + 22*n**2 + 24*n + 8) + 8*A*a*b*x**2* x**n/(6*n**3 + 22*n**2 + 24*n + 8) + 3*A*b**2*n**2*x**2*x**(2*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 8*A*b**2*n*x**2*x**(2*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 4*A*b**2*x**2*x**(2*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 6*B*a**2*n**2 *x**2*x**n/(6*n**3 + 22*n**2 + 24*n + 8) + 10*B*a**2*n*x**2*x**n/(6*n**3 + 22*n**2 + 24*n + 8) + 4*B*a**2*x**2*x**n/(6*n**3 + 22*n**2 + 24*n + 8) + 6*B*a*b*n**2*x**2*x**(2*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 16*B*a*b*n*x**2 *x**(2*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 8*B*a*b*x**2*x**(2*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 2*B*b**2*n**2*x**2*x**(3*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 6*B*b**2*n*x**2*x**(3*n)/(6*n**3 + 22*n**2 + 24*n + 8) + 4*B*b**2* x**2*x**(3*n)/(6*n**3 + 22*n**2 + 24*n + 8), True))
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.22 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {1}{2} \, A a^{2} x^{2} + \frac {B b^{2} x^{3 \, n + 2}}{3 \, n + 2} + \frac {B a b x^{2 \, n + 2}}{n + 1} + \frac {A b^{2} x^{2 \, n + 2}}{2 \, {\left (n + 1\right )}} + \frac {B a^{2} x^{n + 2}}{n + 2} + \frac {2 \, A a b x^{n + 2}}{n + 2} \] Input:
integrate(x*(a+b*x^n)^2*(A+B*x^n),x, algorithm="maxima")
Output:
1/2*A*a^2*x^2 + B*b^2*x^(3*n + 2)/(3*n + 2) + B*a*b*x^(2*n + 2)/(n + 1) + 1/2*A*b^2*x^(2*n + 2)/(n + 1) + B*a^2*x^(n + 2)/(n + 2) + 2*A*a*b*x^(n + 2 )/(n + 2)
Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (72) = 144\).
Time = 0.13 (sec) , antiderivative size = 275, normalized size of antiderivative = 3.62 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2 \, B b^{2} n^{2} x^{2} x^{3 \, n} + 6 \, B a b n^{2} x^{2} x^{2 \, n} + 3 \, A b^{2} n^{2} x^{2} x^{2 \, n} + 6 \, B a^{2} n^{2} x^{2} x^{n} + 12 \, A a b n^{2} x^{2} x^{n} + 3 \, A a^{2} n^{3} x^{2} + 6 \, B b^{2} n x^{2} x^{3 \, n} + 16 \, B a b n x^{2} x^{2 \, n} + 8 \, A b^{2} n x^{2} x^{2 \, n} + 10 \, B a^{2} n x^{2} x^{n} + 20 \, A a b n x^{2} x^{n} + 11 \, A a^{2} n^{2} x^{2} + 4 \, B b^{2} x^{2} x^{3 \, n} + 8 \, B a b x^{2} x^{2 \, n} + 4 \, A b^{2} x^{2} x^{2 \, n} + 4 \, B a^{2} x^{2} x^{n} + 8 \, A a b x^{2} x^{n} + 12 \, A a^{2} n x^{2} + 4 \, A a^{2} x^{2}}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \] Input:
integrate(x*(a+b*x^n)^2*(A+B*x^n),x, algorithm="giac")
Output:
1/2*(2*B*b^2*n^2*x^2*x^(3*n) + 6*B*a*b*n^2*x^2*x^(2*n) + 3*A*b^2*n^2*x^2*x ^(2*n) + 6*B*a^2*n^2*x^2*x^n + 12*A*a*b*n^2*x^2*x^n + 3*A*a^2*n^3*x^2 + 6* B*b^2*n*x^2*x^(3*n) + 16*B*a*b*n*x^2*x^(2*n) + 8*A*b^2*n*x^2*x^(2*n) + 10* B*a^2*n*x^2*x^n + 20*A*a*b*n*x^2*x^n + 11*A*a^2*n^2*x^2 + 4*B*b^2*x^2*x^(3 *n) + 8*B*a*b*x^2*x^(2*n) + 4*A*b^2*x^2*x^(2*n) + 4*B*a^2*x^2*x^n + 8*A*a* b*x^2*x^n + 12*A*a^2*n*x^2 + 4*A*a^2*x^2)/(3*n^3 + 11*n^2 + 12*n + 4)
Time = 3.68 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {A\,a^2\,x^2}{2}+\frac {x^{2\,n}\,x^2\,\left (A\,b^2+2\,B\,a\,b\right )}{2\,n+2}+\frac {x^n\,x^2\,\left (B\,a^2+2\,A\,b\,a\right )}{n+2}+\frac {B\,b^2\,x^{3\,n}\,x^2}{3\,n+2} \] Input:
int(x*(A + B*x^n)*(a + b*x^n)^2,x)
Output:
(A*a^2*x^2)/2 + (x^(2*n)*x^2*(A*b^2 + 2*B*a*b))/(2*n + 2) + (x^n*x^2*(B*a^ 2 + 2*A*a*b))/(n + 2) + (B*b^2*x^(3*n)*x^2)/(3*n + 2)
Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.99 \[ \int x \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {x^{2} \left (2 x^{3 n} b^{3} n^{2}+6 x^{3 n} b^{3} n +4 x^{3 n} b^{3}+9 x^{2 n} a \,b^{2} n^{2}+24 x^{2 n} a \,b^{2} n +12 x^{2 n} a \,b^{2}+18 x^{n} a^{2} b \,n^{2}+30 x^{n} a^{2} b n +12 x^{n} a^{2} b +3 a^{3} n^{3}+11 a^{3} n^{2}+12 a^{3} n +4 a^{3}\right )}{6 n^{3}+22 n^{2}+24 n +8} \] Input:
int(x*(a+b*x^n)^2*(A+B*x^n),x)
Output:
(x**2*(2*x**(3*n)*b**3*n**2 + 6*x**(3*n)*b**3*n + 4*x**(3*n)*b**3 + 9*x**( 2*n)*a*b**2*n**2 + 24*x**(2*n)*a*b**2*n + 12*x**(2*n)*a*b**2 + 18*x**n*a** 2*b*n**2 + 30*x**n*a**2*b*n + 12*x**n*a**2*b + 3*a**3*n**3 + 11*a**3*n**2 + 12*a**3*n + 4*a**3))/(2*(3*n**3 + 11*n**2 + 12*n + 4))